Find $\dfrac{d}{dx}\left(\dfrac{\sin(x)}{x^2}\right)$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{x\cos(x)-2\sin(x)}{x^3}$ (Choice B) B $\dfrac{\cos(x)}{2x}$ (Choice C) C $\dfrac{x\cos(x)+2\sin(x)}{x}$ (Choice D) D $\dfrac{\cos(x)-2x}{x^4}$
$\dfrac{\sin(x)}{x^2}$ is the quotient of two, more basic, expressions: $\sin(x)$ and $x^2$. Therefore, the derivative of the expression can be found using the quotient rule : $\begin{aligned} \dfrac{d}{dx}\left[\dfrac{u(x)}{v(x)}\right]&=\dfrac{\dfrac{d}{dx}[u(x)]v(x)-u(x)\dfrac{d}{dx}[v(x)]}{[v(x)]^2} \\\\ &=\dfrac{u'(x)v(x)-u(x)v'(x)}{[v(x)]^2} \end{aligned}$ Let's differentiate! $\begin{aligned} &\phantom{=}\dfrac{d}{dx}\left(\dfrac{\sin(x)}{x^2}\right) \\\\ &=\dfrac{\dfrac{d}{dx}(\sin(x))x^2-\sin(x)\dfrac{d}{dx}(x^2)}{(x^2)^2}&&\gray{\text{The quotient rule}} \\\\ &=\dfrac{\cos(x)\cdot x^2-\sin(x)\cdot 2x}{(x^2)^2}&&\gray{\text{Differentiate }\sin(x)\text{ and }x^2} \\\\ &=\dfrac{x\left(x\cos(x)-2\sin(x)\right)}{x^4}&&\gray{\text{Simplify}} \\\\ &=\dfrac{x\cos(x)-2\sin(x)}{x^3}&&\gray{\text{Cancel common factors}} \end{aligned}$ In conclusion, $\dfrac{d}{dx}\left(\dfrac{\sin(x)}{x^2}\right)=\dfrac{x\cos(x)-2\sin(x)}{x^3}$